SOLUTION: Solve the equation 4sin^2(theta) + 2 cos(theta) = 3 for 0 < theta < 2 pi
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Question 1141185: Solve the equation 4sin^2(theta) + 2 cos(theta) = 3 for 0 < theta < 2 pi
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
start with 4 * sin^2(theta) + 2 * cos(theta) = 3
since sin^2(theta) + cos^2(theta) = 1, then sin^2(theta) = 1 - cos^2(theta).
your equation becomes 4 * (1 - cos^2(theta) + 2 * cos(theta) = 3
simplify to get 4 - 4 * cos^2(theta) + 2 * cos(theta) = 3
multiply both sides of this equation by -1 to get -4 + 4 * cos^2(theta) - 2 * cos(theta) = -3
subtract 3 from both sides of this equation to get 4 * cos^2(theta) - 2 * cos(theta) - 1 = 0
factor this quadratic equation to get cos(theta) = 0.80901699437495 or cos(theta) = -0.30901699437495
solve for theta to get theta = 36 degrees or theta = 108 degrees.
you are looking for the answer between 0 and 360 degrees.
cosine is positive in first and fourth quadrant.
therefore 36 degrees and 360 - 36 = 324 degrees should be the angles that get you a cosine of 0.80901699437495.
cosine is negative in the second and third quadrants.
therefore 108 degrees and 252 degrees should be the angles that get you a cosine of -0.30901699437495.
note that the reference angle for 108 degrees is 180 - 108 = 72 degrees.
note also that the equivalent angle in the third quadrant is 180 + 72 degrees = 252 degrees.
confirm by replacing theta in the original equation with 36 degrees and 324 degrees and with 108 degrees and 252 degrees.
the original equation should be true with any of these.
i did the math and confirmed that all angles make the original equation true.
you can also solve this by graphing.
the graph looks like this.
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